Đáp án: $x\in\{-1,-\dfrac{17}{7}\}$
Giải thích các bước giải:
ĐKXĐ: $x\ne 1,-\dfrac53,-2,-3$
Ta có :
$\dfrac{1}{x-1}+\dfrac{6}{3x+5}=\dfrac{2}{x+2}+\dfrac{1}{x+3}$
$\to \dfrac{6}{3x+5}-\dfrac{2}{x+2}=\dfrac{1}{x+3}-\dfrac{1}{x-1}$
$\to \dfrac{6(x+2)-2(3x+5)}{(3x+5)(x+2)}=\dfrac{x-1-(x+3)}{(x+3)(x-1)}$
$\to \dfrac{2}{(3x+5)(x+2)}=\dfrac{-4}{(x+3)(x-1)}$
$\to (x+3)(x-1)=-2(3x+5)(x+2)$
$\to x^2+2x-3=-6x^2-22x-20$
$\to 7x^2+24x+17=0$
$\to (x+1)(7x+17)=0$
$\to x+1=0\to x=-1$ hoặc $7x+17=0\to x=-\dfrac{17}{7}$
$\to x\in\{-1,-\dfrac{17}{7}\}$
